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I am trying to understand the space of all orthogonal tensors, I asked a more general version of this question here but with no solution yet found. I want to look at the simplest case first, namely a $2\times2\times2$ tensor. This means if the $2\times2\times2$ tensor is denoted by $a_{i,j,k}$, then the following three equations must hold:

$$ \sum_{i=1}^2 \sum_{j=1}^2 a_{i,j,1} \overline{a}_{i,j,2} = 0 $$

$$\sum_{i=1}^2 \sum_{k=1}^2 a_{i,1,k} \overline{a}_{i,2,k} = 0 $$

$$\sum_{j=1}^2 \sum_{k=1}^2 a_{1,j,k} \overline{a}_{2,j,k} = 0 $$

Except this time instead of considering the most general case of the $2\times2\times2$ tensor being all complex, I want to consider a simpler version. That is where $a_{1,1,1}$, $a_{2,1,1}$, $a_{1,2,1}$, $a_{1,1,2}$ are positive reals and the rest of the entries are complex values. How can one characterize the space of solutions to these three equations? I do believe that the solution space should be 6 dimensional.

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  • $\begingroup$ Why are you chosing only the values of $a_{1,1,1},a_{2,1,1},a_{1,2,1},a_{1,1,2}$ to be positive reals? And when you say dimensional, do you mean topological, algebraic, etc...? $\endgroup$ Commented Sep 6 at 14:23
  • $\begingroup$ @PedroJuanSoto I am investigating the space of unique core tensors in the HOSVD decomposition. I have concluded that requiring the values $a_{1,1,1}$, $a_{2,1,1}$, $a_{1,2,1}$ to be positive reals makes it somewhat unique. And by dimension I mean topological. $\endgroup$
    – jujumumu
    Commented Sep 7 at 22:35

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